Question

If ${x^2} = y + z$ , ${y^2} = z + x$ , ${z^2} = x + y$ , then the value of $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$ is,
A) – 1
B) 1
C) 2
D) 4

Answer 1

Hint:
We can take the expression which we need to find the value of. Then we can multiply and divide with x, y and z in the respective terms. Then we can substitute the given equations. Then we can simplify and cancel the common terms to get the required solution.

Complete step by step solution:
We need to find the value of $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$.
Let $K = \dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$
We know that the value of a fraction remains the same when multiplied with the same number on both numerator and denominator. So, we can multiply and divide the \[{1^{st}}\] term with x.
$ \Rightarrow K = \dfrac{x}{{x\left( {x + 1} \right)}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$
Similarly, we can multiply and divide the \[{2^{nd}}\] term with y and multiply and divide the 3rd term with z.
$ \Rightarrow K = \dfrac{x}{{x\left( {x + 1} \right)}} + \dfrac{y}{{y\left( {y + 1} \right)}} + \dfrac{z}{{z\left( {z + 1} \right)}}$
On expanding the bracket, we get,
$ \Rightarrow K = \dfrac{x}{{{x^2} + x}} + \dfrac{y}{{{y^2} + y}} + \dfrac{z}{{{z^2} + z}}$
We are given that ${x^2} = y + z$ , ${y^2} = z + x$ and ${z^2} = x + y$ . So, on substituting these values in the denominators of the above equation, we get
$ \Rightarrow K = \dfrac{x}{{y + z + x}} + \dfrac{y}{{x + z + y}} + \dfrac{z}{{x + y + z}}$
As the denominators are equal, we can add the numerators. So, we get
$ \Rightarrow K = \dfrac{{x + y + z}}{{x + y + z}}$
As the numerators and denominators are equal, we can cancel them.
$ \Rightarrow K = 1$
So, we have
$ \Rightarrow \dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}} = 1$
Therefore, the required value of the expression is 1.

So, the correct answer is option B.

Note:
Alternate method to solve this problem is given by
 We need to find the value of $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$ .
Let $K = \dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$
We write the numerator of the \[{1^{st}}\] term as $1 = 1 + x - x$ and similarly the \[{2^{nd}}\] term and \[{3^{rd}}\] term in terms of y and z respectively.
$ \Rightarrow K = \dfrac{{1 + x - x}}{{x + 1}} + \dfrac{{1 + y - y}}{{y + 1}} + \dfrac{{1 + z - z}}{{z + 1}}$
On splitting the numerator, we get
$ \Rightarrow K = \dfrac{{1 + x}}{{x + 1}} - \dfrac{x}{{x + 1}} + \dfrac{{1 + y}}{{y + 1}} - \dfrac{y}{{y + 1}} + \dfrac{{1 + z}}{{z + 1}} - \dfrac{z}{{z + 1}}$
As the numerators and denominators are equal, we can cancel them.
$ \Rightarrow K = 1 - \dfrac{x}{{x + 1}} + 1 - \dfrac{y}{{y + 1}} + 1 - \dfrac{z}{{z + 1}}$
Now we can multiply and divide with x, y and z in the respective terms.
$ \Rightarrow K = 3 - \left( {\dfrac{{{x^2}}}{{{x^2} + x}} + \dfrac{{{y^2}}}{{{y^2} + y}} + \dfrac{{{z^2}}}{{{z^2} + z}}} \right)$
We are given that ${x^2} = y + z$ , ${y^2} = z + x$ and ${z^2} = x + y$ . So, on substituting these values in the denominators of the above equation, we get
$ \Rightarrow K = 3 - \left( {\dfrac{{y + z}}{{y + z + x}} + \dfrac{{x + z}}{{x + z + y}} + \dfrac{{x + y}}{{x + y + z}}} \right)$
As the denominators are equal, we can add the numerators. So, we get
$ \Rightarrow K = 3 - \dfrac{{2\left( {x + y + z} \right)}}{{x + y + z}}$
As the numerators and denominators are equal, we can cancel them.
$ \Rightarrow K = 3 - 2$
On simplification, we get
$ \Rightarrow K = 1$
 So, we have
$ \Rightarrow \dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}} = 1$
Therefore, the required value of the expression is 1.